# The upper bound estimate of the number of integer points on elliptic curves ${y}^{2}={x}^{3}+{p}^{2r}x$

- Jin Zhang
^{1}and - Xiaoxue Li
^{2}Email author

**2014**:104

https://doi.org/10.1186/1029-242X-2014-104

© Zhang and Li; licensee Springer. 2014

**Received: **22 January 2014

**Accepted: **20 February 2014

**Published: **4 March 2014

## Abstract

Let *p* be a fixed prime and *r* be a fixed positive integer. Further let $N({p}^{2r})$ denote the number of pairs of integer points $(x,\pm y)$ on the elliptic curve $E:{y}^{2}={x}^{3}+{p}^{2r}x$ with $y>0$. Using some properties of Diophantine equations, we give a sharper upper bound estimate for $N({p}^{2r})$. That is, we prove that $N({p}^{2r})\le 1$, except with $N({17}^{2(2s+1)})=2$, where *s* is a nonnegative integer.

**MSC:**11G05, 11Y50.

### Keywords

elliptic curve integer point Diophantine equation## 1 Introduction

*p*be a fixed prime and

*k*be a fixed positive integer. Recently, the integer points $(x,y)$ on the elliptic curve

*k*. Then (1.1) can be rewritten as

where *r* is a positive integer.

*n*, let

Using some properties of Diophantine equations, we give a sharper upper bound estimate for $N({p}^{2r})$, the number of pairs of non-trivial integer points $(x,\pm y)$ on (1.2). That is, we shall prove the following results.

**Theorem 1.1**

*All non*-

*trivial integer points on*(1.2)

*are given as follows*.

- (i)
$p=u({2}^{m})$, $r=2s+1$, $(x,\pm y)=({p}^{2s}{v}^{2}({2}^{m}),\pm {p}^{3s}v({2}^{m})({v}^{2}({2}^{m})+1))$,

*where**m*,*s**are nonnegative integers*. - (ii)$p\equiv 1(mod\phantom{\rule{0.25em}{0ex}}8)$, $r=2s+1$, $(x,\pm y)=({p}^{2s+1}{X}^{2},\pm {p}^{3s+2}XY)$,
*where**s**is a nonnegative integer*, $(X,Y)$*is a solution of the equation*${X}^{4}-p{Y}^{2}=1,\phantom{\rule{1em}{0ex}}X,Y\in \mathbb{N}.$(1.5)

**Theorem 1.2** *Let* *p* *be an odd prime*, *r* *be a positive integer*. *Then for any nonnegative integer* *s*, *we have* $N({p}^{2r})\le 1$, *except with* $N({17}^{2(2s+1)})=2$. *Moreover*, *if* $p\not\equiv 1(mod\phantom{\rule{0.25em}{0ex}}8)$, *then* $N({p}^{2r})=0$, *except with* $N({3}^{2(2s+1)})=1$.

## 2 Lemmas

**Lemma 2.1** ([[4], Theorem 244])

*Every solution*$(u,v)$

*of the equation*

*can be expressed as* $(u,v)=(u(n),v(n))$, *where* *n* *is a positive integer*.

**Lemma 2.2** *If* $p=u(n)$, *then* $n={2}^{m}$, *where* *m* *is a nonnegative integer*.

*Proof* Assume that *n* has an odd divisor *d* with $d>1$. Then we have either $u(1)|u(n)$ and $1<u(1)<u(n)$ or $u(n/d)|u(n)$ and $1<u(n/d)<u(n)$. Therefore, since *p* is a prime, it is impossible. Thus, we get $n={2}^{m}$. The lemma is proved. □

Any fixed positive integer *a* can be uniquely expressed as $a=b{c}^{2}$, where *b*, *c* are positive integers with *b* is square free. Then *b* is called the quadratfrei of *a* and denoted by $Q(a)$.

**Lemma 2.3** *For any positive integer* *m*, *we have* $3|Q(v({2}^{m}))$.

*Proof*By (1.3) and (1.4), we get

Since $(2/3)=-1$, where $(2/3)$ is the Legendre symbol, we see from (2.3) that $3\nmid u({2}^{i})$ for $i\ge 1$. Therefore, since $u(1)=3$, by (2.2), we obtain $3\parallel v({2}^{m})$. It implies that $3|Q(v({2}^{m}))$. The lemma is proved. □

*D*be a non-square positive integer. It is a well known fact that if the equation

*l*, let

Then $(U,V)=(U(l),V(l))$ ($l=1,3,\dots $) are all solutions of (2.4).

**Lemma 2.4** ([[5], Theorem 1])

*The equation*

*has at most one solution* $(X,Y)$. *Moreover*, *if the solution* $(X,Y)$ *exists*, *then* $({X}^{2},Y)=(U(l),V(l))$, *where* $l=Q({U}_{1})$.

**Lemma 2.5** ([[5], Theorem 3])

*If* $3|Q({U}_{1})$, *then* (2.5) *has no solutions* $(X,Y)$.

**Lemma 2.6** *If* $p=u({2}^{m})$, *where* *m* *is a positive integer with* $m>1$, *then* (1.5) *has no solutions* $(X,Y)$.

*Proof*Since $p=u({2}^{m})$ with $m>1$, by (2.3), we have

has solution $(U,V)$ and its fundamental solution is $({U}_{1},{V}_{1})=(2v({2}^{m-1}),1)$. Further, since $m-1\ge 1$, by Lemma 2.3, we have $3|Q(v({2}^{m-1}))$. Hence, we get $3|Q({U}_{1})=Q(2v({2}^{m-1}))$. Therefore, by Lemma 2.5, the lemma is proved. □

**Lemma 2.7** ([6])

*The equation*

*has no solutions* $(X,Y,n)$.

**Lemma 2.8**

*The equation*

*has only the solutions* $(p,X,Y,n)=(u({2}^{m}),{v}^{2}({2}^{m})+1,v({2}^{m}),1)$, *where* *m* *is a nonnegative integer*.

*Proof* Assume that $(p,X,Y,n)$ is a solution of (2.9). If $p=2$, since $gcd(X,Y)=1$, then we have $2\nmid XY$, $gcd(X+{Y}^{2},X-{Y}^{2})=2$, $X+{Y}^{2}={2}^{2n-1}$, $X-{Y}^{2}=2$ and ${Y}^{2}={2}^{2n-2}-1$. But since ${Y}^{2}+1$ is not a square, it is impossible.

*p*is an odd prime, then we have $gcd(X+{Y}^{2},X-{Y}^{2})=1$, and by (2.9), we get $X+{Y}^{2}={p}^{2n}$, $X-{Y}^{2}=1$,

Further, by Lemma 2.2, we see from the first equality of (2.13) that $n={2}^{m}$. Thus, by (2.10) and (2.13), the lemma is proved. □

## 3 Proof of Theorem 1.1

*x*can be expressed as

Therefore, by (3.1), (3.4), and (3.6), the integer points of type (i) are given.

It implies that $(X,Y)=(f,g)$ is a solution of (1.5). Therefore, by (3.1) and (3.8), we obtain the integer points of type (ii).

Since $p\nmid z({p}^{2t-2r}{z}^{2}+1)$ and $gcd(z,{p}^{2t-2r}{z}^{2}+1)=1$, we see from (3.10) that ${p}^{2t-2r}{z}^{2}+1$ is a square, a contradiction.

To sum up, the theorem is proved.

## 4 Proof of Theorem 1.2

It implies that the theorem is true for $p\not\equiv 1(mod\phantom{\rule{0.25em}{0ex}}8)$.

and $N({17}^{2(2s+1)})=2$. However, by Lemma 2.6, if $p=u({2}^{m})$ with $m>1$, then ${N}_{2}=0$. Therefore, by (4.2), if $p\equiv 1(mod\phantom{\rule{0.25em}{0ex}}8)$, then $N({p}^{2r})\le 1$, except with $N({17}^{2(2s+1)})=2$. The theorem is proved.

## Declarations

### Acknowledgements

The authors would like to thank the referees for their very helpful and detailed comments, which have significantly improved the presentation of this paper. This work is supported by the P.E.D. (2013JK0573) and N.S.F. (11371291) of P.R. China.

## Authors’ Affiliations

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